As shown in the figure, a block of mass $\sqrt{3}\, kg$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3 \sqrt{3}}$. The critical force to be applied on the vertical surface as shown at an angle $60^{\circ}$ with horizontal such that it does not move, will be $3 x$. The value of $3x$ will be
$\left[ g =10 m / s ^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]$
JEE MAIN 2021, Difficult
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$F \cos 60^{\circ}=\mu N$ or $\frac{ F }{2}=\frac{1}{3 \sqrt{3}} N$ $\ldots$ (1)
$\& N=\sin 60^{\circ}+\sqrt{3} g$ $\ldots(2)$
From equation
$(1)\;and\;(2)$
$\frac{F}{2}=\frac{1}{3 \sqrt{3}}\left(\frac{F \sqrt{3}}{2}+\sqrt{3} g\right)$
$\Rightarrow F = g =10$ Newton $=3 x$
So $x=\frac{10}{3}=3.33$
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